Question

(13+4b -3c )^2 evaluate​

Answer 1

$\huge\bf\blue{Hey\:Mate}$

We will solve this by using the identity,

${x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2xz= {(x + y + z)}^{2}$

Here we go...

$= {(13 + 4b - 3c)}^{2} \\ = {(13)}^{2} + {(4b)}^{2} + {( - 3c)}^{2} + 2(13)(4b) + 2(4b)( - 3c) + 2(13)( - 3c) \\ = 169 + 16 {b}^{2} + 9 {c}^{2} + 104b - 24bc - 78c$

@ItsMysteriousGirl✒️

Answer 2

Answer:

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